fractional binomial limit sum

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fractional binomial limit sum

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#1 h Apr 19, 2025, 9:51 PM

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this was given to me by a friend

$\lim_{n \to \infty} \sum_{k=1}^{n}{\frac{1}{\binom{n}{k}}}$

a nice solution using sandwich is
$\frac{1}{n} + \frac{1}{n} + 1 + \frac{n-3}{\binom{n}{2}} \ge \frac{1}{n} + \sum_{k=2}^{n-2}{\frac{1}{\binom{n}{k}}}+ \frac{1}{n} + 1 \ge \frac{1}{n} + + \frac{1}{n} + 1$

therefore $\lim_{n \to \infty} \sum_{k=1}^{n}{\frac{1}{\binom{n}{k}}}$ = $1$

ALSO ANOTHER SOLUTION WHICH I WAS THINKING OF WITHOUT SANDWICH BUT I CANT COMPLETE WAS TO USE THE GAMMA FUNCTION

we know

$B(x, y) = \int_0^1 t^{x - 1} (1 - t)^{y - 1} \, dt$

$B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)}$

and $\Gamma(n) = (n-1)!$ for integers,

$\frac{1}{\binom{n}{k}}$ = $\frac{k! (n-k)!}{n!}$

therefore from the gamma function we get

$(n+1) \int_{0}^{1} x^k (1-x)^{n-k} dx$ = $\frac{1}{\binom{n}{k}}$ = $\frac{k! (n-k)!}{n!}$
$\Rightarrow$ $\lim_{n \to \infty} (n+1) \int_{0}^{1} \sum_{k=1}^{n} x^k (1-x)^{n-k} dx$ $=\lim_{n \to \infty} \sum_{k=1}^{n}{\frac{1}{\binom{n}{k}}}$

somehow im supposed to show that

$\lim_{n \to \infty} (n+1) \int_{0}^{1} \sum_{k=1}^{n} x^k (1-x)^{n-k} dx$ $= 1$

all i could observe was if we do L'hopital (which i hate to do as much as you do)

i get $\frac{ \int_{0}^{1} \sum_{k=1}^{n} x^k (1-x)^{n-k} dx}{1/n+1}$

now since $x \in (0,1)$ , as $n \to \infty$ the $(1-x)^{n-k} \to 0$ which gets us the $\frac{0}{0}$ form therefore L'hopital came to my mind , which might be a completely wrong intuition, anyway what should i do to find that limit

:noo:

:noo:

:pilot:

This post has been edited 3 times. Last edited by Levieee, Apr 20, 2025, 10:14 AM

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KAME06

#2 h Apr 19, 2025, 11:49 PM

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Check Putnam 1958 B1, you might find something useful there

This post has been edited 1 time. Last edited by KAME06, Apr 19, 2025, 11:51 PM

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#3 h Apr 20, 2025, 7:51 AM

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btw the limit must be at least $2$ (and turns out it is precisely $2$ ) as the first and the last term are both $1$ i.e, $1/\binom{n}{0}+1/\binom{n}{n}=2$ .

This post has been edited 1 time. Last edited by anudeep, Apr 20, 2025, 7:52 AM

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#4 h Apr 20, 2025, 9:44 AM

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anudeep wrote:

btw the limit must be at least $2$ (and turns out it is precisely $2$ ) as the first and the last term are both $1$ i.e, $1/\binom{n}{0}+1/\binom{n}{n}=2$ .

oh my bad edited the question

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